I am interested in criteria that guarantee that a Borel probability measure has compact support. I outline two below and I am hoping to gather more as answers (if they exist). ---------- The first sufficient condition I know for a Borel probability measure $\mu$ (say) on $\mathbb{R}$ which has all moments of order $k$ for any $k\ge 1$ is as follows. If there exists $C>0$ such that for any $k\ge 1$, $$\int |x|^k d\mu(x)\le C R^k,$$ then $$\operatorname{supp}\mu \subseteq \{|x|\le R\}.$$ The proof uses Chebyshev's inequality. Let $\lambda>R$. Then, $$\mu\left\{x\in \mathbb{R}:|x|>\lambda\right\}\le \frac{1}{\lambda^k}\int |x|^kd\mu(x)\le C \left(\frac{R}{\lambda}\right)^k\to 0$$ if we let $k\to\infty.$ Of course, the converse also holds. ---------- The second condition is the Paley-Wiener result given in this [answer][1]. ---------- Unlike in the linked answers above, I am interested in merely *sufficient* conditions guaranteeing the compactness of the support with the hope that there are more results of this kind in addition to the ones I quoted above. [1]: https://mathoverflow.net/questions/129688/a-sufficient-condition-for-a-probability-measure-to-have-compact-support#:~:text=In%20order%20to%20avoid%20circularity,axis%20outside%20of%20this%20interval.